A company determines that its total profit is given by the function 

Upper P left parenthesis x right parenthesis equals negative 2 x squared plus 402 x minus 4180.P(x)=−2x2+402x−4180. 

The company makes a profit for those nonnegative values of x for which 

Upper P left parenthesis x right parenthesis greater than 0.P(x)>0. 

The company loses money for those nonnegative values of x for which 

Upper P left parenthesis x right parenthesis less than 0.P(x)<0. 

Find the values of x for which the company makes a profit and loses money.

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Part 1

The company makes a profit on the interval 

left parenthesis 11 comma 190 right parenthesis(11,190).

​(Type your answer in interval​ notation.)

Part 2

The company loses money on the interval 

enter your response here. 

​(Type your answer in interval​ notation.)

To find the intervals where the company makes a profit or loses money, we need to analyze the given quadratic profit function:

\[
P(x) = -2x^2 + 402x - 4180
\]

### Part 1: Finding when the company makes a profit

The company makes a profit when \(P(x) > 0\). Since \(P(x)\) is a quadratic function (a parabola that opens downward due to the negative coefficient of \(x^2\)), we can solve for the values of \(x\) by finding where \(P(x) = 0\), which gives the boundaries of the interval where \(P(x) > 0\).

1. Set \(P(x) = 0\) to find the roots:
   \[
   -2x^2 + 402x - 4180 = 0
   \]
   
2. Solve this quadratic equation using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \(a = -2\), \(b = 402\), and \(c = -4180\).

   Substituting these values:
   \[
   x = \frac{-402 \pm \sqrt{402^2 - 4(-2)(-4180)}}{2(-2)}
   \]
   \[
   x = \frac{-402 \pm \sqrt{161604 - 33440}}{-4}
   \]
   \[
   x = \frac{-402 \pm \sqrt{128164}}{-4}
   \]
   \[
   x = \frac{-402 \pm 357.92}{-4}
   \]

   Now, solve for both roots:
   \[
   x_1 = \frac{-402 + 357.92}{-4} = \frac{-44.08}{-4} = 11.02 \approx 11
   \]
   \[
   x_2 = \frac{-402 - 357.92}{-4} = \frac{-759.92}{-4} = 189.98 \approx 190
   \]

So, the roots are \(x_1 = 11\) and \(x_2 = 190\).

Since the parabola opens downward, the profit is positive between these two roots. Therefore, the company makes a profit for:

\[
x \in (11, 190)
\]

### Part 2: Finding when the company loses money

The company loses money when \(P(x) < 0\). This happens outside the interval where the company makes a profit, meaning for values of \(x\) less than 11 and greater than 190.

Thus, the company loses money for:

\[
x \in (-\infty, 11) \cup (190, \infty)
\]

### Final Answer:

- **Profit Interval**: \((11, 190)\)
- **Loss Interval**: \((-\infty, 11) \cup (190, \infty)\)

Would you like further clarification on solving quadratic equations, or more details on the intervals?

Here are 5 related questions:
1. What are the general steps to solve a quadratic inequality like \(P(x) > 0\)?
2. How does the shape of a parabola determine where the function is positive or negative?
3. Can you explain how the discriminant helps determine the nature of the roots of a quadratic equation?
4. How do you use interval notation to describe the solution to inequalities?
5. What real-world applications might involve determining when a company makes a profit or loss?

**Tip**: Always check the direction the parabola opens (upward or downward) when interpreting the sign of the quadratic function in different intervals.

A company determines that its total profit is given by the function P(x) = −2x^2 + 402x − 4180. The company makes a profit for those nonnegative values of x for which P(x) > 0. The company loses money for those nonnegative values of x for which P(x) < 0. Find the values of x for which the company makes a profit and loses money.

Discover how to find the intervals where a company makes a profit or loses money using the quadratic profit function P(x) = −2x^2 + 402x − 4180. This problem covers key concepts in algebra, such as quadratic inequalities and the quadratic formula. Suitable for students in Grades 9-12.

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